To solve the trigonometric equation $\sin^2(x) - 1 = 0$ for $0 \leq x \leq 2\pi$, follow these steps:

1. Rewrite the Equation:

   The given equation is: $\sin^2(x) - 1 = 0$

   Add $1$ to both sides to isolate $\sin^2(x)$: $\sin^2(x) = 1$

2. Solve for $\sin(x)$:

   To find the values of $x$, take the square root of both sides: $\sin(x) = \pm 1$

3. Find the Angles Where $\sin(x) = 1$ and $\sin(x) = -1$:

   • For $\sin(x) = 1$, this occurs when $x = \frac{\pi}{2}$.
   • For $\sin(x) = -1$, this occurs when $x = \frac{3\pi}{2}$.

4. List the Solutions:

   Therefore, the values of $x$ that satisfy the equation within the interval $0 \leq x \leq 2\pi$ are: $x = \frac{\pi}{2}, \quad x = \frac{3\pi}{2}$

So, the solution to the equation $\sin^2(x) - 1 = 0$ within the interval $0 \leq x \leq 2\pi$ is:

$x = \frac{\pi}{2}, \quad \frac{3\pi}{2}$

Would you like more details or have any questions about this solution?

Related Questions:

1. How would you solve $\cos^2(x) - 1 = 0$ for $0 \leq x \leq 2\pi$?
2. What are the general solutions to $\sin^2(x) = 1$ over all real numbers?
3. How do you find solutions to $\sin(x) = 0$ within the interval $0 \leq x \leq 2\pi$?
4. Can you solve $\sin(2x) = 1$ for $0 \leq x \leq 2\pi$?
5. What are the solutions to $\cos(x) = 1$ for $0 \leq x \leq 2\pi$?

Tip:

To solve trigonometric equations, it's helpful to recall the unit circle values where sine and cosine have specific known values, like $0, \pm 1, \pm \frac{\sqrt{2}}{2}$.